
The standard Gibbs free energy $\Delta \text{G}{}^\circ $ is related to the equilibrium constant ${{\text{K}}_{P}}$ as:
$\begin{align}
& \text{A}\text{. }{{\text{K}}_{P}}\text{ = -RTlog}\vartriangle \text{G}{}^\circ \\
& \text{B}\text{. }{{\text{K}}_{P}}\text{ = }{{\left( \text{e}/\text{RT} \right)}^{\vartriangle \text{G}{}^\circ }} \\
& \text{C}\text{. }{{\text{K}}_{P}}\text{ = -}\vartriangle \text{G}{}^\circ /\text{RT} \\
& \text{D}\text{. }{{\text{K}}_{P}}\text{ = }{{\text{e}}^{\text{-RT/}\vartriangle \text{G}{}^\circ }} \\
\end{align}$
Answer
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Hint: Gibbs free energy is that thermodynamic quantity, the decrease in whose value during a process is equal to the maximum possible useful work that can be obtained from the system. An equilibrium constant tells us about the reaction progress and to what extent will it proceed.
Complete step by step answer:
-Firstly, we will see the relation between the free energy change of the reaction i.e. the relation between Gibbs free energy (\[\Delta \text{G}\]) and standard Gibbs free energy (\[\Delta \text{G}{}^\circ \]) i.e. \[\Delta \text{G = }\Delta \text{G}{}^\circ \text{ + RTlnQ }....\text{(1)}\]
-Here, Q is the reaction quotient.
-When equilibrium state is attained, then the value of \[\Delta \text{G}\]becomes zero and Q becomes equal to k.
-So, equation (1) becomes:
\[\text{0 = }\Delta \text{G}{}^\circ \text{ + RTlnk}\]
\[\Delta \text{G}{}^\circ \text{ = -RTlnk }.....\text{(2)}\]
-Now, equation (2) can also be written as:
\[\text{ k = }{{\text{e}}^{\Delta \text{G}{}^\circ \text{/RT}}}\]
-The reaction is also known as Von't Hoff Reaction Isotherm.
-The value of \[\Delta \text{G}{}^\circ \]for the endothermic process is positive because the value of \[\Delta \text{H}{}^\circ \] is positive and large.
-Whereas the value of \[\Delta \text{G}{}^\circ \] for the exothermic reaction is negative because the value of \[\Delta \text{H}{}^\circ \]is negative and large.
So, the correct answer is “Option D”.
Note: The value of the reaction quotient is equal to the \[{{\text{Q}}_{P}}\] if the reactant and product are present in the gaseous state whereas the value reaction quotient is equal to \[{{\text{Q}}_{\text{c}}}\] when the reactant and product are present in the same solution. The relation between standard Gibbs free energy, enthalpy change and entropy change can as: \[\Delta \text{G}{}^\circ \text{ = }\Delta \text{H}{}^\circ \text{ - T}\Delta \text{S}{}^\circ \]
Complete step by step answer:
-Firstly, we will see the relation between the free energy change of the reaction i.e. the relation between Gibbs free energy (\[\Delta \text{G}\]) and standard Gibbs free energy (\[\Delta \text{G}{}^\circ \]) i.e. \[\Delta \text{G = }\Delta \text{G}{}^\circ \text{ + RTlnQ }....\text{(1)}\]
-Here, Q is the reaction quotient.
-When equilibrium state is attained, then the value of \[\Delta \text{G}\]becomes zero and Q becomes equal to k.
-So, equation (1) becomes:
\[\text{0 = }\Delta \text{G}{}^\circ \text{ + RTlnk}\]
\[\Delta \text{G}{}^\circ \text{ = -RTlnk }.....\text{(2)}\]
-Now, equation (2) can also be written as:
\[\text{ k = }{{\text{e}}^{\Delta \text{G}{}^\circ \text{/RT}}}\]
-The reaction is also known as Von't Hoff Reaction Isotherm.
-The value of \[\Delta \text{G}{}^\circ \]for the endothermic process is positive because the value of \[\Delta \text{H}{}^\circ \] is positive and large.
-Whereas the value of \[\Delta \text{G}{}^\circ \] for the exothermic reaction is negative because the value of \[\Delta \text{H}{}^\circ \]is negative and large.
So, the correct answer is “Option D”.
Note: The value of the reaction quotient is equal to the \[{{\text{Q}}_{P}}\] if the reactant and product are present in the gaseous state whereas the value reaction quotient is equal to \[{{\text{Q}}_{\text{c}}}\] when the reactant and product are present in the same solution. The relation between standard Gibbs free energy, enthalpy change and entropy change can as: \[\Delta \text{G}{}^\circ \text{ = }\Delta \text{H}{}^\circ \text{ - T}\Delta \text{S}{}^\circ \]
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